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\cl {{\bf Space Curves of Constant Curvature }
 \footnote"* "{\verysmall 
 This file is from the 3D-XplorMath project.  Please see: \hfill\break
   \phantom{http://} http://3D-XplorMath.org/}}

\vskip5pt
\noindent
\hbox{ 
\vbox{\hsize=0.5\hsize \phantom{.}\hskip-5mm
\includegraphics[width=1.3in] {Constant_Curvature.png}}
\vbox{\hsize=0.45\hsize \noindent
2 - 11 Torus Knot of \lf
constant curvature.
\lf\lf
See also: \lf About  Spherical Curves\lf
}}
\vskip-3pt
\noindent
{\it  Definition via Differential Equations}.
Space Curves that 3DXM can exhibit are mostly given in terms of explicit formulas
or explicit geometric constructions. The differential geometric treatment of curves
starts from such examples and defines geometric properties, i.e., properties which
do not change when the curve is transformed by an isometry (= distance preserving
map, also called a rigid motion) of Euclidean space $\BR^3$. The most important such 
properties are the curvature function $\kappa$ and the torsion function $\tau$. Once 
they have been defined one proves the {\it Fundamental Theorem of Space Curves},
which states that for any given continuous functions $\kappa,\tau$ there is a space
curve with these curvature and torsion functions, and, that this curve is uniquely 
determined up to a rigid motion.
\vskip2pt
To define curvature, observe that at each point of a parametrized space curve
$c(t)$ there is a parametrized  circle $\gamma(t)$ with \lf\phantom{.}\hskip1cm
$c(t_0)=\gamma(t_0),\dot c(t_0)=\dot\gamma(t_0), \ddot c(t_0)=\ddot\gamma(t_0).$
\lf
This circle -- which may degenerate to a straight line -- is called the {\it osculating
circle at $t_0$}, its radius is called {\it curvature radius  at $t_0$} and the inverse
of the radius is called {\it the curvature  at $t_0$}, $\kappa(t_0)$. 
The computation of curvature is simpler if the curve is parametrized by arc
length, i.e. if the length of all tangent vectors is one, $|\dot c(t)|=1$. One gets
$\kappa(t) = |\ddot c(t)| $. Check this for the circle $c(t):= r\cdot (\cos(t/r),\ \sin(t/r) )$.
The most common way to proceed is to assume
that $\kappa(t) >0$. This allows one to define the {\it Frenet basis} along the curve:
\vskip-16pt
$$\eqalign{
&e_1(t) := \dot c(t), \cr
&e_2(t) := \ddot c(t)/\kappa(t), \cr
&e_3(t):= e_1(t)\times e_2(t).
}$$
\vskip-6pt \noindent
The Frenet basis defines three curves $t \mapsto  e_j(t)$ on the unit sphere.
To emphasize the fact that $e_j(t)$ are to be considered as vectors, not as
points, one calls the length of their derivative, $|\dot e_j(t)|$, {\it angular velocity}
or {\it rotation speed} and not just velocity. For example, the formula
$\ddot c(t) = \kappa(t) e_2(t)$ says that $\kappa(t)$ is the rotation speed of $\dot c(t)$.
Next, we get from $\dot e_1(t) \sim e_2(t)$ that the derivative of $e_3(t)$ is proportional
to $e_2(t)$. This proportionality factor, the rotation speed of $e_3(t)$, is called the 
torsion function $\tau(t)$ of the curve $c(t)$. In formulas:
$\tau(t) := \langle \dot e_3(t),\ e_2(t) \rangle $. 
\medskip\goodbreak
\noindent
Now one changes the point of view and considers the two functions 
$\kappa, \tau$ as given. This turns the equations that were originally
definitions of  $\kappa$ and $ \tau$ into differential equations, the famous
{\it Frenet-Serret Equations: } 
\goodbreak  \phantom{.} \vskip-0.7cm
$$\eqalign{
&\dot e_1(t) = \hskip1cm\kappa(t)\cdot e_2(t), \cr
&\dot e_2(t) = -\kappa(t)\cdot e_1(t) - \tau(t)\cdot e_3(t), \cr
&\dot e_3(t)= \hskip1cm\tau(t)\cdot e_2(t),
}$$
or, with \hskip2mm$\vec\omega(t) := -\tau(t)\cdot e_1(t) + \kappa(t)\cdot e_3(t)$,
\phantom{.} \vskip-0.6cm
$$\eqalign{  \dot e_j(t) &= \vec\omega(t) \times e_j(t).  \hskip3.cm\cr
  \phantom{.} \hskip-0.6cm  \hbox{ Finally } \hskip5mm \dot c(t) &= e_1(t).
    }$$
 \vskip-6pt\noindent
For given continuous functions $\kappa,\tau$ these differential equations
have --- for given orthonormal initial values --- unique orthonormal solutions
$\{e_1(t), e_2(t), e_3(t) \}$. The curve $c(t) := \int ^t e_1(s)ds$ is then parametrized
by arc length and has the given curvature functions $\kappa, \tau$.
\smallskip
\noindent 
The simplest curves in the plane, straight lines and circles, have constant curvature.
One may wonder what constant curvature curves look like in $\BR^3$.
In 3DXM we illustrate the use of the Frenet-Serret equations by showing the
following family of constant curvature curves: 


$ \kappa(t) := aa,$

$\tau(t):= bb + cc\cdot\sin(t) + dd\cdot\sin(2t) + ee\cdot\sin(3t)$. \lf
\vskip -6pt \noindent
The function $\tau$ is, if $bb=0$, skew symmetric at its zeros at $0$ and $\pi$. 
This implies that the  solution curves are symmetric with respect to the normal 
planes at these points. From
this it follows that we can get {\it closed nonplanar curves of constant curvature}
easily: the only requirement is that the angle of the normal planes at $c(0)$ and
$c(\pi)$ has to be a rational multiple of $\pi$. {\it Every $bb=0$ one-parameter family 
of examples in 3DXM therefore contains many closed examples} --- select in
the Animation Menu the default morph. 
\vskip3pt\noindent
In the less symmetric case $bb \ne 0$ (but $dd=0$) the function $\tau$ is even 
at the maxima and the minima, at $t=\pi/2,\ t=3\pi/2$, and this implies that $180^\circ$
rotation around $e_2$ at these points is also a symmetry of solution curves. This
can be used to find more closed curves by solving 2-parameter problems as follows:
For every value of $aa, bb$ use $cc$ to make the distance between the normals $0$.
Now change $aa$ or $bb$ slowly (continuing to use $cc$ for keeping the distance
between the normals $0$) and observe how the angle between the symmetry 
normals varies. If this angle hits a value $2k/n\cdot\pi$ then $n$ copies of the computed
piece fit together to a smoothly closed curve. \lf
If one has selected `Constant Curvature' in the Menu 
`Space Curves' then there is in the Action Menu an entry `Other Closed Curves'.
It opens a submenu where one can select first $bb=0$ examples which are also hit
by the default morph. Then there are $bb \ne 0$ embedded examples, some of them
knotted. Moreover, the 11-2-knot has nonvanishing torsion and strongly resembles
a torus knot. This is no coincidence since one can find constant curvature curves on
tori by solving a second order ODE, and it is again a 2-parameter problem to close
these up. -- The example `like 6 helices' looks in another way as one would imagine
constant curvature curves: made up of left winding and right winding pieces of helices.
\lf
Do not miss selecting `Show Osculating Circles $\&$ Evolute'. The constant radius of the 
osculating circles shows the constant curvature and the rotating motion of the radius
shows size and sign of the torsion.
\vskip3pt%\hrule
\noindent
In 3DXM one can choose in the Action Menu {`Parallel Frame'}.This frame is designed 
to rotate as little as possible along the curve, in $\BR^3$.
This property is more obvious when one looks at the torus knots than at the constant 
curvature curves. For further details see curves of {\it constant torsion}. The main 
advantage of these parallel frames is that they neither make it neccessary to assume 
more than continuity of the second derivative $\ddot{c}$, nor that 
$\kappa > 0$ everywhere, even straight lines are not exceptional curves if one 
works with these frames. Their differential equation is also simple:
\vskip-10pt
$$\eqalign{
&\hbox{\it Frenet-Serret Equations for Parallel Frames: } \cr
&\dot e_1(t) := \hskip1cm a(t)\cdot e_2(t) +b(t)\cdot e_3(t), \cr
&\dot e_2(t) :=  -a(t)\cdot e_1(t), \cr
&\dot e_3(t):=  -b(t)\cdot e_1(t).
}$$ 
With an anti-derivative $T(t)$ of the torsion
$\tau(t) = T'(t)$ we can of course write the two-dimensional curvature vector 
$(a(t), b(t))$ in terms of $\kappa(t), \tau(t)$, namely:
$$
(a(t), b(t)) := \kappa(t)\big(\cos(T(t)), \sin(T(t)) \big).
$$

\smallskip\noindent
H.K.
 
\bye


\lf\phantom{.}\hskip1cm  Note $\kappa(t) = \sqrt{a^2(t) + b^2(t)}$.